On Saturday 06 September 2003 06:01, Darryl wrote:
> I have some php code that pulls from the mysql database. Here it is:
>
> <?php
> mysql_connect("wildcat.osborneindustries.com", "webuser",
> "webpass");
> $mymonth = date('n');
> $cyear = date('Y');
> $query = "SELECT name,hdate FROM emp2 where month(hdate)=$mymonth
> order by hdate";
> $result = mysql_db_query("iweb", $query);
>
> if ($result) {
> echo "<table align=center border=0 cellspacing=5 >";
> while ($r = mysql_fetch_array($result)) {
> $name = $r["name"];
> $hyear = date('Y',$r["hdate"]);
> $timein = $cyear - $hyear;
> if ($timein > 0) {
> echo
> "<tr><td>$name</t><td>$timein</td><td>$cyear</td><td>$hyear</td></tr>";}
> }
> echo "</table>";
> } else {
> echo "No data.";
> }
> mysql_free_result($result); ?>
>
> I'm trying to figure out years of employment based on hiredate. Its not
> working as expected. $timein is always
> 1969. Date in the database is YYYY-MM-DD. What have I screwed up ?
date() in php expects a unix timestamp. Dates in MySQL are NOT stored in unix
timestamp format.
You can perform the calculations directly in MySQL
mysql manual > Tutorial Introduction > Date Calculations
--
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